exercise 5.3.2. if $phi(x,y,z)=3x^{2}y - y^{3}z^{2}$, find $\nablaphi$ at the point $(1, - 2, - 1)$

Explanation:

Step1: Recall gradient formula

The gradient of a scalar - valued function $\phi(x,y,z)$ is given by $\nabla\phi=\left(\frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y},\frac{\partial\phi}{\partial z}\right)$.

Step2: Calculate $\frac{\partial\phi}{\partial x}$

Differentiate $\phi(x,y,z) = 3x^{2}y - y^{3}z^{2}$ with respect to $x$ treating $y$ and $z$ as constants. Using the power - rule $\frac{\partial}{\partial x}(ax^{n})=nax^{n - 1}$, we have $\frac{\partial\phi}{\partial x}=6xy$.

Step3: Calculate $\frac{\partial\phi}{\partial y}$

Differentiate $\phi(x,y,z)$ with respect to $y$ treating $x$ and $z$ as constants. $\frac{\partial\phi}{\partial y}=3x^{2}-3y^{2}z^{2}$.

Step4: Calculate $\frac{\partial\phi}{\partial z}$

Differentiate $\phi(x,y,z)$ with respect to $z$ treating $x$ and $y$ as constants. $\frac{\partial\phi}{\partial z}=-2y^{3}z$.

Step5: Evaluate at the point $(1, - 1,-1)$

For $\frac{\partial\phi}{\partial x}$ at $(1,-1,-1)$: $\frac{\partial\phi}{\partial x}\big|{(1,-1,-1)}=6\times1\times(-1)=-6$. For $\frac{\partial\phi}{\partial y}$ at $(1,-1,-1)$: $\frac{\partial\phi}{\partial y}\big|{(1,-1,-1)}=3\times1^{2}-3\times(-1)^{2}\times(-1)^{2}=3 - 3=0$. For $\frac{\partial\phi}{\partial z}$ at $(1,-1,-1)$: $\frac{\partial\phi}{\partial z}\big|_{(1,-1,-1)}=-2\times(-1)^{3}\times(-1)=-2$.

Answer:

$(-6,0,-2)$